wiz-icon
MyQuestionIcon
MyQuestionIcon
15
You visited us 15 times! Enjoying our articles? Unlock Full Access!
Question

Instantaneous velocity of a particle moving in a straight line is given as v=(4+4t) m/s, for the first 5 sec of its motion, there after velocity becomes constant. Find the acceleration and displacement of the particle at time t=3.0 sec.

A
23 m/s2, 35.50m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
23 m/s2, 25.85m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
5 m/s2, 25.85m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10 m/s2, 15.85m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 23 m/s2, 25.85m
Given,
velocity of particle for the first 5 sec, v=(4+4t) m/s

As we know that acceleration of a particle is time derivative of its instantaneous velocity.a=dvdt a=d(4+4t)dt

a=2t

And at time t=3.0 sec, acceleration is

a=23 m/s2

To find displacement: We know that velocity is the time derivative of displacement, we have to integrate as follows

v=dxdt=4+4t

dx=(4+4t)dt

Integrating the above expression from time t=0 to t=3.0 sec, we get the displacement, which varies from xi=0 to xf=x, for these time instants.

xf=xxi=0dx=t=3t=0(4+4t)dt

[x0]=[4t+83t3/2]30

x=[12+13.85]

x25.85 m

Hence, option (b) is the correct answer.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon